Guess the Biggest?

Let $k=2017$ ; then we have:

$\begin{cases} a = \sqrt{k+1} + \sqrt{k-1} \\ b = \sqrt{k+2} + \sqrt{k-2} \\ c = \sqrt{k+3} + \sqrt{k-3} \end{cases} \implies \begin{cases} a^2 = 2k + 2\sqrt{k^2-1} \\ b^2 = 2k + 2\sqrt{k^2-4} \\ c^2 = 2k + 2\sqrt{k^2-9} \end{cases}$

Therefore, $a^2 > b^2 > c^2$ $\implies a > b > c$ , hence $\boxed a$ is the biggest.

My solution is similar to that of Jian Hau Chooi but it is slightly different .

Let us assume that $x = 2017$ $\begin{aligned} a = \sqrt{x + 1} + \sqrt{x - 1} \\ b = \sqrt{x + 2} + \sqrt{x - 2} \\ c = \sqrt{x + 3} + \sqrt{x - 3} \\ \end{aligned}$ Now, squaring on both the sides we get : $\begin{aligned} a^2 = x + 1 + x - 1 + 2\sqrt{x^2 - 1} = {\color{#20A900}2x + 2}\sqrt{\color{#3D99F6}x^2 - 1} \\ b^2 = x + 2 + x - 2 + 2\sqrt{x^2 - 4} = {\color{#20A900}2x + 2}\sqrt{\color{#E81990}x^2 - 4} \\ c^2 = x + 3 + x - 3 + 2\sqrt{x^2 - 9} = {\color{#20A900}2x + 2}\sqrt{\color{#D61F06}x^2 - 9} \\ \end{aligned}$ Now, again square root on both the sides : $\begin{aligned} a = \sqrt{{\color{#20A900}2x + 2}\sqrt{\color{#3D99F6}x^2 - 1}} \\ b = \sqrt{{\color{#20A900}2x + 2}\sqrt{\color{#E81990}x^2 - 4}} \\ c = \sqrt{{\color{#20A900}2x + 2}\sqrt{\color{#D61F06}x^2 - 9}} \\ \end{aligned}$ Now, the $\color{#20A900}green~part$ is same for $a,b,c$ but they differ in the inside square root part. We also know that if the number in the inner square root is less than the number as a whole is larger . $\begin{aligned} \sqrt{\color{#3D99F6}x^2 - 1} > \sqrt{\color{#E81990}x^2 - 4} > \sqrt{\color{#D61F06}x^2 - 9} \\ \implies a > b > c \quad \quad \quad \quad \\ \end{aligned}$

$\therefore a$ is the largest number.

Notice that $\dfrac{a -\sqrt{2016}}{b-\sqrt {2015}} = \dfrac{\sqrt{2018}}{\sqrt{2019}} < 1 \implies a -\sqrt{2016} < b -\sqrt{2015} \\\implies a - b = \sqrt{2016} -\sqrt{2015} > 0 \implies \boxed{a > b}$ Further, $\dfrac{b - \sqrt{2015} }{c - \sqrt{2014}} =\dfrac{\sqrt{2019}}{\sqrt{2020}} < 1 \implies b - \sqrt{2015} < c - \sqrt{2014} \\ b-c = \sqrt{2015} -\sqrt{2014} >0 \implies \boxed{b> c}$ Eventually, $a > b , b > c \implies a> c \\\boxed{\therefore a > b > c}$ Thus, $a$ is greatest among them.

For $n \geq k \geq 0$ , $\large \begin{aligned} r &= \sqrt{n+k}+\sqrt{n-k} \\ r^2 &= 2n+2\sqrt{n^2-k^2} \ \ \rightarrow\ \ \ r\propto\frac{1}{k} \end{aligned}$

By taking $n=2017$ : $\begin{aligned} k &= 1,\ \ a=\sqrt{2018}+\sqrt{2016} \\ k &= 2,\ \ b=\sqrt{2019}+\sqrt{2015} \\ k &= 3,\ \ c=\sqrt{2020}+\sqrt{2014} \\ \end{aligned}$

Hence, we will found that $a>b>c$ , and $\boxed a$ will be the biggest.